| Topic: |
Science > Physics |
| User: |
"peter" |
| Date: |
22 Sep 2005 08:33:07 AM |
| Object: |
diffusion conundrum |
Can anyone throw light on this?
Consider a closed, hollow cylinder completely filled with a fluid.The
cylinder is placed at the centre of a turntable which is rotated at a
constant rate. After some time all the fluid reaches the angular
velocity of the turntable-there is steady state. Except for the random
thermal jiggling of the molecules of the fluid it is as if the cylinder
were a solid. (There will be a pressure gradient (and if the fluid is
compressible there will be a density gradient) along the radius of the
cylinder, but I don't think this affects the argument.)
Consider an imaginary cylindrical surface, coaxial with the cylinder
axis, within the rotating fluid. Thermal fluctuations guarantee that
there will be a two way traffic of molecules across this surface.
Since the average density remains constant at each point or radius in
the fluid then the flow rate of molecules inward must equal the flow
rate outward. But the inward diffusing molecules have a greater
'drift' velocity (w * r) than the outward ones and consequently
there is a continuous transfer of momentum and no steady state.
Is diffusion suppressed in a rotating fluid?
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| User: "Androcles Androcles@ MyPlace.org" |
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| Title: Re: diffusion conundrum |
22 Sep 2005 09:10:06 AM |
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"peter" <topeter@bigpond.net.au> wrote in message
news:1127395986.966240.249520@g14g2000cwa.googlegroups.com...
| Can anyone throw light on this?
|
| Consider a closed, hollow cylinder completely filled with a fluid.
Yeah, ok, a beer can.
| The
| cylinder is placed at the centre of a turntable which is rotated at a
| constant rate.
You'd put a beer can on an antique gramophone?
I hope this is a thought experiment only.
| After some time all the fluid reaches the angular
| velocity of the turntable-there is steady state.
Same effect when you shake the can... too much froth.
| Except for the random
| thermal jiggling of the molecules of the fluid it is as if the
cylinder
| were a solid.
Why not put it in the freezer?
That will open the can if you let it go solid.
| (There will be a pressure gradient (and if the fluid is
| compressible there will be a density gradient) along the radius of the
| cylinder, but I don't think this affects the argument.)
Try opening the beer can, there'll be pressure all right.
| Consider an imaginary cylindrical surface, coaxial with the cylinder
| axis, within the rotating fluid.
Now you want a toilet roll core in the can? Yeuch....
It's your beer, not mine.
| Thermal fluctuations guarantee that
| there will be a two way traffic of molecules across this surface.
| Since the average density remains constant at each point or radius in
| the fluid then the flow rate of molecules inward must equal the flow
| rate outward. But the inward diffusing molecules have a greater
| 'drift' velocity (w * r) than the outward ones and consequently
| there is a continuous transfer of momentum and no steady state.
|
| Is diffusion suppressed in a rotating fluid?
Yeah... the diffused bubbles go to the middle instead of the top.
Pass the unspun beer over, there's a good chap.
Androcles
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| User: "Puppet_Sock" |
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| Title: Re: diffusion conundrum |
22 Sep 2005 11:44:52 AM |
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peter wrote:
Can anyone throw light on this?
Consider a closed, hollow cylinder completely filled with a fluid.The
cylinder is placed at the centre of a turntable which is rotated at a
constant rate. After some time all the fluid reaches the angular
velocity of the turntable-there is steady state. Except for the random
thermal jiggling of the molecules of the fluid it is as if the cylinder
were a solid. (There will be a pressure gradient (and if the fluid is
compressible there will be a density gradient) along the radius of the
cylinder, but I don't think this affects the argument.)
Consider an imaginary cylindrical surface, coaxial with the cylinder
axis, within the rotating fluid. Thermal fluctuations guarantee that
there will be a two way traffic of molecules across this surface.
Since the average density remains constant at each point or radius in
the fluid then the flow rate of molecules inward must equal the flow
rate outward. But the inward diffusing molecules have a greater
'drift' velocity (w * r) than the outward ones and consequently
there is a continuous transfer of momentum and no steady state.
Is diffusion suppressed in a rotating fluid?
Maarten gave a technical answer, which is of course the complete
and correct way to go. But, for those of us who find it hard to
recall our fluid mechanics (that was 1980 for me, after all) here
is a less technical though less accurate answer. (Do read
Maarten's answer.)
Yes, there is a w * r effect. And it's balanced by the pressure
gradient. It's where the pressure gradient comes from. What,
after all, is a pressure gradient? Suppose that the density
changes very little. How can the pressure change? What does
a higher pressure fluid at (nearly) the same density look like
on a microscopic scale? Well, in some fashion, it has to be
giving either more, or harder, impulse through molecule motion.
So, the w * r term is pitching the molecules outward harder,
and the higher pressure is pitching them back harder. And when
these things exactly balance you get equilibrium.
By the by, something similar must be going on in a column of
fluid in a gravity field. The pressure differential has to be
balanced by the gravity differential. And if you think about
particles moving down getting extra energy from gravity, and
particles moving up losing it, then clearly something very
similar is going on. You reach equilibrium when this is exactly
balanced by the pressure gradient.
Socks
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| User: "AJW" |
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| Title: Re: diffusion conundrum |
22 Sep 2005 01:04:20 PM |
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It's almost trivial to suggest this model, but think of any kind of
gradient: temperature, concentration, force, whatever. The 'stuff'' in
which there's a gradient will try to slide down the gradient field, and
the speed of its sliding -- we might call it diffusion -- has to do
with the rate of change of concentration with respect to distance. When
something is in equilibrum, there has to be some counter-force. A neat
example is a pH electrode. When there's a difference in H ions, the
ions want to migrate through the glass, say from left to right, and
will until an opposing electrical force resists them. Actually, it just
forces as many ions to go to the left as the gradient makes slide to
the right.
It gets to be even better: if you think in terms of temperature, for
example, and you know the thermal capacity and conductivity of the
material in which the gradient exists, you can begin thinking in terms
of how the temperature profile changes w/r/t time. That kind of
transient analysis is quite a lot of fun.
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| User: "Maarten van Reeuwijk" |
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| Title: Re: diffusion conundrum |
22 Sep 2005 10:22:07 AM |
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peter wrote:
Consider an imaginary cylindrical surface, coaxial with the cylinder
axis, within the rotating fluid. Thermal fluctuations guarantee that
there will be a two way traffic of molecules across this surface.
Since the average density remains constant at each point or radius in
the fluid then the flow rate of molecules inward must equal the flow
rate outward. But the inward diffusing molecules have a greater
'drift' velocity (w * r) than the outward ones and consequently
there is a continuous transfer of momentum and no steady state.
Interesting argument, and your reasoning is correct. There is a diffusive
transport, but this is counterbalanced by another force due to the
rotation. The problem is best described in cylindrical coordinates,
(http://scienceworld.wolfram.com/physics/Navier-StokesEquations.html), and
for this particular example (with ur=0, uz=0, d/dz=0 and d/dtheta=0), the
momentum equations reduce to:
(conservation of momentum in r-direction)
-utheta^2/r=-dp/dr (1)
(conservation of momentum in theta-direction)
-1/r(d/dr(r d utheta / dr)) - utheta/r^2 = 0 (2)
Eqn (1) is the balance you described between centripetal forcing and
pressure. Eqn (2) is about your question, with the first term being the
diffusion you were speaking about (see the inside term r d utheta / dr),
and the second term arising from the transformation to cylindrical
coordinates.
Normally, eqn (2) is neglected. Assume that the typical velocity U = omega *
D, with u the angular velocity and D the diameter of the cylinder. The
centripetal force will be O(omega^2 * D) and the diffusive forces will be
O(omega/D). So the ratio centripetal/diffusion = omega^2 D which will
normally much larger than 1, so it is safe to neglect (2).
Is diffusion suppressed in a rotating fluid?
In summary, no, it is not suppressed, it goes on just like usual.
HTH, Maarten
PS. It is better to ask questions re fluid mechanics in sci.mech.fluids,
that increases the chance of useful answers substantially.
PPS. Rotating fluids, see also a previous thread:
http://www.usenet.com/newsgroups/
sci.physics.computational.fluid-dynamics/msg00085.html
--
===================================================================
Maarten van Reeuwijk dept. of Multiscale Physics
Phd student Faculty of Applied Sciences
maarten.ws.tn.tudelft.nl Delft University of Technology
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| User: "Maarten van Reeuwijk" |
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| Title: Re: diffusion conundrum |
22 Sep 2005 10:32:51 AM |
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The centripetal force will be O(omega^2 * D) and the diffusive forces will
be O(omega/D).
Typo: the diffusive forces will be O(nu omega / D) with nu the dynamic
viscosity.
So the ratio centripetal/diffusion = omega^2 D/nu which will
normally much larger than 1, so it is safe to neglect (2).
--
===================================================================
Maarten van Reeuwijk dept. of Multiscale Physics
Phd student Faculty of Applied Sciences
maarten.ws.tn.tudelft.nl Delft University of Technology
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| User: "peter" |
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| Title: Re: diffusion conundrum |
28 Sep 2005 07:58:27 AM |
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Maarten
It is possible that I have misunderstood you-I am not an expert.
However.
1. the expression "is diffusion suppressed etc." was only half serious.
I cannot imagine it.
2. if you have the time and inclination could you state your answer to
my problem in non-mathematical language
3. your expression for the tangential acceleration is identically zero
-1/r(d/dr(r d utheta / dr)) - utheta/r^2 = 0 (2)
i.e. (omega/r) - (omega/r)
= 0
4. in the physical situation that I posed there is no relative motion
between layers of fluid so why does viscosity appear in your
explanation
5. even if the influence of diffusion compared to centripetal force is
small how does this affect the argument? If there is a continuous
transfer of momentum between adjacent layers of rotating fluid via
diffusion then the stable motion with constant angular velocity (omega)
that we observe should not happen.
6. Note that the argument that I am proposing relies on the mechanism
that is used to explain the origin of viscosity in gases (not liquids
as far as I know)
7. congratulations on your English. But then you are Dutch-natural
masters of English
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| User: "Maarten van Reeuwijk" |
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| Title: Re: diffusion conundrum |
29 Sep 2005 08:18:40 AM |
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peter wrote:
Maarten
It is possible that I have misunderstood you-I am not an expert.
However.
1. the expression "is diffusion suppressed etc." was only half serious.
I cannot imagine it.
2. if you have the time and inclination could you state your answer to
my problem in non-mathematical language
3. your expression for the tangential acceleration is identically zero
-1/r(d/dr(r d utheta / dr)) - utheta/r^2 = 0 (2)
i.e. (omega/r) - (omega/r)
= 0
4. in the physical situation that I posed there is no relative motion
between layers of fluid so why does viscosity appear in your
explanation
5. even if the influence of diffusion compared to centripetal force is
small how does this affect the argument? If there is a continuous
transfer of momentum between adjacent layers of rotating fluid via
diffusion then the stable motion with constant angular velocity (omega)
that we observe should not happen.
6. Note that the argument that I am proposing relies on the mechanism
that is used to explain the origin of viscosity in gases (not liquids
as far as I know)
Imagine your cylindrical object filled with a liquid. If there was no
viscosity, this object could spin as fast as possible and the fluid would
stay at rest. It is viscosity that is responsible for redistributing
momentum wherever there are differences in velocity. This phenomenon is
represented by eqn 2 but now with the time-derivative included
d utheta / dt -1/r(d/dr(r d utheta / dr)) - utheta/r^2 = 0 (2)
If utheta=0 (fluid at rest), then this equation becomes
d utheta / dt = 1/r(d/dr(r d utheta / dr)) + 0
and because the rhs is locally very large, the fluid will accelerate until
an equilibrium between viscosity and the body force utheta/r^2 is reached.
The fact that you find zero upon substitution of utheta = omega r in (2)
only means that you can congratulate yourself: you have found a solution to
this complicated-looking ODE!
With respect to point 5: the steady state is between the viscosity and the
body force utheta/r^2 (which is a result of the coordinate transformation),
so the equilibrium and thus steady state is between these two forces.
7. congratulations on your English. But then you are Dutch-natural
masters of English
Thank you, although not as fluent as you: I had to look your 'conundrum' and
'ponder' up :).
HTH, Maarten
--
===================================================================
Maarten van Reeuwijk dept. of Multiscale Physics
Phd student Faculty of Applied Sciences
maarten.ws.tn.tudelft.nl Delft University of Technology
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| User: "peter" |
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| Title: Re: diffusion conundrum |
23 Sep 2005 08:22:07 AM |
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thank you for your thoughtful comments. I will ponder and reply
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